I am trying to add a 0 infront of dates 1 through 9 as the dates are listed in the dropdown menu. Here is my code, i thought using d would add leading zeros but doesn't seem to be working. I do not have much php experience so this is a longshot... thank you in advanced!
<?PHP
FUNCTION DateSelector($inName, $useDate=0)
{
/* create array so we can name months */
$monthName = ARRAY(1=> "January", "February", "March",
"April", "May", "June", "July", "August",
"September", "October", "November", "December");
/* if date invalid or not supplied, use current time */
IF($useDate == 0)
{
$useDate = TIME();
}
/* make month selector */
ECHO "<SELECT NAME=" . $inName . "month>\n";
FOR($currentMonth = 1; $currentMonth <= 12; $currentMonth++)
{
ECHO "<OPTION VALUE=\"";
ECHO INTVAL($currentMonth);
ECHO "\"";
IF(INTVAL(DATE( "m", $useDate))==$currentMonth)
{
ECHO " SELECTED";
}
ECHO ">" . $monthName[$currentMonth] . "\n";
}
ECHO "</SELECT>";
/* make day selector */
ECHO "<SELECT NAME=" . $inName . "day>\n";
FOR($currentDay=1; $currentDay <= 31; $currentDay++)
{
ECHO "<OPTION VALUE=\"$currentDay\"";
IF(INTVAL(DATE( "d", $useDate))==$currentDay)
{
ECHO " SELECTED";
}
ECHO ">$currentDay\n";
}
ECHO "</SELECT>";
/* make year selector */
ECHO "<SELECT NAME=" . $inName . "year>\n";
$startYear = DATE( "Y", $useDate);
FOR($currentYear = $startYear - 0; $currentYear <= $startYear+2;$currentYear++)
{
ECHO "<OPTION VALUE=\"$currentYear\"";
IF(DATE( "Y", $useDate)==$currentYear)
{
ECHO " SELECTED";
}
ECHO ">$currentYear\n";
}
ECHO "</SELECT>";
}
?>
If i understand you correctly, you need this:
$day = 1;
echo str_pad($day, 2, 0, STR_PAD_LEFT);
You can replace:
ECHO INTVAL($currentMonth);
with:
printf("%02s", $currentMonth);
$day_with_leading_zeroes = sprintf("%02d", $day);
instead of:
ECHO ">$currentDay\n";
you can type:
echo ">".($currentDay<10 ? "0" : "").$currentDay."\n";
$date =4
$month = 6
$year = 2013
if want display above in this format. 04/06/2013
printf('%02d/%02d/%04d', $date, $month, $year);
$date =14
$month = 12
$year = 2013
if want display above in this format. 14/12/2013
printf('%02d/%02d/%04d', $date, $month, $year);
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